Reduction by LiAlH₄ of hydrolyzed product of an ester gives:

1. Two alcohols

2. Two aldehyde

3. One acid and one alcohol

4. Two acids

Intermediates formed during the reaction of ${\mathrm{RCNH}}_2$
$$ $$ $$ $\left|\right|$
$$ $$ $$ $\mathrm{O}$ with ${\mathrm{Br}}_2$ and KOH are:

1. RNHCOBr and RNCO

2. RCONHBr and RNCO

3. RNH-Br and RCONHBr 

4. RCON${\mathrm{Br}}_2$

In the following reaction, product 'P' is:

1. RCH₂OH

2. RCOOH

3. RCHO

4. RCH₃

In a set of the given reactions, acetic acid yielded a product C.

\(\mathrm{CH_3COOH+PCl_5\rightarrow A\xrightarrow[Anh.AlCl_3]{C_6H_6}B\xrightarrow[ether]{C_2H_5MgBr}C}\)
Product C would be:

1.	CH3CH(OH)C2H5
2.	CH3COC6H5
3.	CH3CH(OH)C6H5
4.

When m-chlorobenzaldehyde is treated with 50% KOH solution, the product(s) obtained is (are):

1.
2.
3.
4.